Solved Paper 2019

Solved Paper 2019 Paper 1 Section 1 (Maximum Marks : 12) This section contains FOUR (04) questions. Each question has FOUR options. ONLY ONE of these four options is the correct answer. For each question, choose the option corresponding to the correct answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : + 3 If ONLY the correct options is chosen. Zero Marks : 0 If none of the options is chosen. (i.e. the question is unanswered) Negative Marks : - 1 In all other cases. l l l l 1. The correct order of acid strength of the following carboxylic acids is O O I OH II H H OH H O III MeO IV H3C OH OH O (a) III > II > I > IV (b) I > II > III > IV (c) II > I > IV > III (d) I > III > II > IV 4. Molar conductivity (L m) of aqueous solution of sodium stearate, which behaves as a strong electrolyte, is recorded at varying concentrations (C) of sodium stearate. Which one of the following plots provides the correct representation of micelle formation in the solution? (critical micelle concentration (CMC) is marked with an arrow in the figures) Lm Lm (a) (b) CMC 2. The green colour produced in the borax bead test of a chromium (III) salt is due to (a) Cr 2 O 3 (c) Cr(BO 2 )3 ÖC ÖC (b) CrB (d) Cr2 (B 4O 7 )3 3. Calamine, malachite, magnetite and cryolite, respectively, are (a) ZnCO 3 , CuCO 3 , Fe 2O 3 , Na 3 AlF6 CMC Lm CMC (c) Lm CMC (d) (b) ZnSO 4 , CuCO 3 , Fe 2O 3 , AlF3 (c) ZnSO 4 , Cu(OH)2 , Fe 3O 4 , Na 3 AlF6 (d) ZnCO 3 , CuCO 3 × Cu(OH)2 , Fe 3O 4 , Na 3 AlF6 ÖC ÖC Section 2 (Maximum Marks : 32) l l l l l This section contains EIGHT (08) questions. Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is (are) correct options(s). For each question, choose the options(s) corresponding to (all) the correct answer(s). Answer to each question will be evaluated according to the following marking scheme: Full Marks : + 4 If only (all) the correct option(s) is (are) chosen. Partial Marks : + 3 If all the four options are correct but ONLY three options are chosen. Partial Marks : + 2 If three or more options are correct but ONLY two options are chosen, both of which are correct and both of which are correct. Partial Marks : + 1 If two or more options are correct but ONLY one option is chosen and it is a correct option; Zero Marks : 0 If none of the option is chosen (i.e., the question is unanswered); Negative Marks : - 1 In all other cases. For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct answers, then choosing ONLY (A), (B) and (D) will get + 4 marks choosing ONLY (A) and (B) will get + 2 marks choosing ONLY (A) and (D) will get + 2 marks choosing ONLY (B) and (D) will get + 2 marks choosing ONLY (A) will get + 1marks choosing ONLY (B) will get + 1marks choosing ONLY (D) will get + 1marks choosing no option (i.e., the question is unanswered) will get 0 marks; and choosing any other combination of options will get - 1mark. 5. Choose the reaction(s) from the following options, for which the standard enthalpy of reaction is equal to the standard enthalpy of formation. (a) 2C(g ) + 3 H2 (g ) ¾® C 2H6 (g ) (c) x1 will deflect towards negatively charged plate (d) x3 is g-ray 8. Which of the following statement(s) is(are) correct regarding the root mean square speed (U rms) and average translational kinetic energy (E av) of a molecule in a gas at equilibrium? (b) 2H2 (g ) + O 2 (g )¾® 2H2O(l ) 3 (c) O 2 (g ) ¾® O 3 (g ) 2 1 (d) S8 (s ) + O 2 (g ) ¾® SO 2 (g ) 8 (a) Urms is inversely proportional to the square root of its molecular mass (b) Urms is doubled when its temperature is increased four times 6. A tin chloride Q undergoes the following reactions (not balanced) Q + Cl - ¾® X (c) Eav is doubled when its temperature is increased four times (d) Eav at a given temperature does not depend on its molecular mass 9. Choose the correct option(s) for the following set of Q + Me3N ¾® Y Q + CuCl 2 ¾® Z + CuCl reactions. C6H10O X is a monoanion having pyramidal geometry. Both Y and Z are neutral compounds. (i) MeMgBr (ii) H2O T (i) H2, Ni (b) The central atom in Z has one lone pair of electrons (c) The oxidation state of the central atom in Z is + 2 (d) The central atom in X is sp3 hybridised CH3 H C 3 - x1 234 90 Th - x2 ¾¾ ¾® 234 91 Pa (b) x2 is b H 3C Br CH3 Cl S 230 90 Th x1 , x 2 , x 3 and x 4 are particles/radiation emitted by the respective isotopes. The correct option(s) is(are) - U (major) (b) - x3 Z ¾¾ ¾® D Cl ¾¾ ¾® - x4 HBr, benzoyl peroxide Br (a) 234 (a) Z is an isotope of uranium R (ii) Br2, hn (major) (major) ¾¾ ¾® S 20 % H3PO4, 360 K (a) There is a coordinate bond in Y 238 92 U Conc. HCl (major) Choose the correct option(s). 7. In the decay sequence. Q H3C Cl T CH3 CH3 Br (c) S U S U H 3C Br Br (d) U T 10. Which of the following statement(s) is(are) true? (a) The two six-membered cyclic hemiacetal forms of D-(+ ) -glucose are called anomers (b) Oxidation of glucose with bromine water gives glutamic acid (c) Monosaccharides cannot be hydrolysed to given polyhydroxy aldehydes and ketones (d) Hydrolysis of sucrose gives dextrorotatory glucose and laevorotatory fructose 11. Fusion of MnO 2 with KOH in presence of O 2 produces a salt W. Alkaline solution of W upon electrolytic oxidation yields another salt X . The manganese containing ions present in W and X , respectively, are Y and Z. Correct statement(s) is (are) (a) Both Y and Z are coloured and have tetrahedral shape (b) Y is diamagnetic in nature while Z is paramagnetic (c) In both Y and Z, p-bonding occurs between p-orbitals of oxygen and d-orbitals of manganese (d) In aqueous acidic solution, Y undergoes disproportionation reaction to give Z and MnO 2 12. Each of the following options contains a set of four molecules. Identify the option(s) where all four molecules posses permanent dipole moment at room temperature. (a) SO 2 , C 6H5Cl, H2Se, BrF5 (b) BeCl 2 , CO 2 , BCl 3 , CHCl 3 (c) NO 2 , NH3 , POCl 3 , CH3Cl (d) BF3 , O 3 , SF6 , XeF6 Section 3 (Maximum Marks : 18) l l l This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE. Four each question, enter the correct numerical value of the answer using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places. Answer to each question will be evaluated according to the following marking scheme: Full Marks : + 3 If ONLY the correct numerical value is entered. Zero Marks : 0 In all other cases. 13. Among B 2H 6, B 3N 3H 6, N 2O, N 2O 4, H 2S2O 3 and H 2S2O 8, the total number of molecules containing covalent bond between two atoms of the same kind is ............. 14. On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of benzene, its vapour pressure decreases from 650 mmHg to 640 mmHg. The depression of freezing point of benzene (in K) upon addition of the solute is ............ (Given data : Molar mass and the molal freezing point depression constant of benzene are 78 g mol - 1 and 512 . K kg mol - 1 , respectively). at 298 K is 16 . ´ 10 17 . Fe2 + (aq) + S2 - (aq) - FeS (s) When equal volumes of 0 .06 M Fe 2+ (aq) and 0 .2 M S2 - (aq) solutions are mixed, the equilibrium concentration of Fe2 + (aq) is found by Y ´ 10 - 17 M. The value of Y is ........... 17. At 143 K, the reaction of XeF4 with O 2F2 produces a xenon compound Y . The total number of lone pair(s) of electrons present on the whole molecule of Y is ............ 15. Consider the kinetic data given in the following table for the reaction A + B + C ¾® Product Experiment No. 16. For the following reaction, the equilibrium constant K c Rate of [A] [B] [C] reaction (mol dm- 3 ) (mol dm- 3 ) (mol dm- 3 ) (mol dm- 3s - 1 ) 18. Schemes 1 and 2 describe the conversion of P to Q and R to S, respectively. Scheme 3 describes the synthesis of T from Q and S. The total number of Br atoms in a molecule of T is .............. Scheme 1 1 0.2 0.1 0.1 6.0 ´ 10 - 5 2 0.2 0.2 0.1 6.0 ´ 10 - 5 3 0.2 0.1 0.2 12 . ´ 10 - 4 4 0.3 0.1 0.1 9.0 ´ 10 - 5 The rate of the reaction for [ A] = 0 .15 mol dm -3 to be Y ´ 10 - 5 mol dm - 3s - 1 . The value of Y is ............ (i) Br2(excess), H2O (ii) NaNO2, HCl, 273 K (iii) CuCN/KCN (iv) H3O+, D (v) SOCl2, pyridine P Scheme 2 , [ B] = 0 .25 mol dm - 3 and [C ] = 0 .15 mol dm - 3 is found NH2 (i) Oleum (ii) NaOH, D (iii) H+ (iv) Br2, CS2, 273 K R Scheme 3 S (i) NaOH, D (ii) Q T (major) Q (major) S (major) Paper 2 Section 1 (Maximum Marks : 32) l l l l l This section contains EIGHT (08) questions. Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four option(s) is (are) correct options(s). For each question, choose the correct options(s) to answer the question. Answer to each question will be evaluated according to the following marking scheme: Full Marks : + 4 If only (all) the correct option(s) is (are) chosen. Partial Marks : + 3 If all the four options are correct but ONLY three options are chosen. Partial Marks : + 2 If three or more options are correct but ONLY two options are chosen, both of which are correct options. Partial Marks : + 1 If two or more options are correct but ONLY one option is chosen and it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : - 2 In all other cases. For example: in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct answer, then choosing ONLY (A), (B) and (D) will get +4 marks; choosing ONLY (A) and (B) will get +2 marks; choosing ONLY (A) and (D) will get +2 marks; choosing ONLY (B) and (D) will get +2 marks; choosing ONLY (A) will get +1mark; choosing ONLY (B) will get +1mark; choosing ONLY (D) will get +1mark; choosing no option (i.e. the question is unanswered) will get 0 marks; and choosing any other combination of options will -1mark. 1. Choose the correct option(s) for the following reaction 2. Choose the correct option(s) that give(s) an aromatic compound as the major product. sequence CHO Hg2+, (i) dil. H2SO4 (ii) AgNO3, NH4OH (iii) Zn-Hg, conc. HCl MeO Q Br (i) SOCl2 pyridine NaOEt (a) (ii) AlCl 3 Zn-Hg conc. HCl R S + Cl2 (excess) (b) Consider Q, R and S as major products. (c) Br H3C (i) Alc. KOH (ii) NaNH2 (iii) Red hot iron tube, 873 K Br (a) MeO UV, 500 K MeO R S O (d) NaOMe CO2H (b) MeO 3. The ground state energy of hydrogen atom is MeO Q R (c) MeO MeO R O S OH CO2H (d) MeO MeO Q S O -13.6 eV. Consider an electronic state Y of He+ whose energy, azimuthal quantum number and magnetic quantum number are -3.4 eV, 2 and 0, respectively. Which of the following statement(s) is(are) true for the state Y? (a) It is a 4d state (b) The nuclear charge experienced by the electron in this state is less than 2e, where e is the magnitude of the electronic charge (c) It has 2 angular nodes (d) It has 3 radial nodes 4. Consider the following reactions (unbalanced). Zn + Hot conc. H 2SO 4 ¾® G + R + X Zn + conc. NaOH ¾® T + Q G + H 2S + NH 4OH ¾® Z (a precipitate) + X + Y Choose the correct option(s). (a) The oxidation state of Zn in T is +1 (b) R is a V-shaped molecule (c) Bond order of Q is 1 in its ground state (d) Z is dirty white in colour 5. With reference to aqua-regia, choose the correct option(s). (a) Aqua-regia is prepared by mixing conc. HCl and conc. HNO 3 in 3 : 1 (v / v ) ratio (b) The yellow colour of aqua-regia is due to the presence of NOCl and Cl 2 (c) Reaction of gold with aqua-regia produces an anion having Au in +3 oxidation state (d) Reaction of gold with aqua regia produces NO 2 in the absence of air (b) Natural rubber is polyisoprene containing trans alkene units (c) Cellulose has only a-D-glucose units that are joined by glycosidic linkages (d) Nylon-6 has amide linkages 7. The cyanide process of gold extraction involves leaching out gold from its ore with CN - in the presence of Q in water to form R. Subsequently, R is treated with T to obtain Au and Z. Choose the correct option(s). (a) Q is O 2 (b) Z is [Zn(CN)4 ]2- (c) T is Zn (d) R is [Au(CN)4 ]- 8. Which of the following reactions produce(s) propane as a major product? (a) H3C Cl Zn, dil. HCl (b) H3C Br Zn (c) H3C COONa (d) H3C COONa + H2O Br NaOH, CaO, D 6. Choose the correct option(s) from the following. (a) Teflon is prepared by heating tetrafluoroethene in presence of a persulphate catalyst at high pressure Electrolysis Section 2 (Maximum Marks : 18 l l l This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value of the answer using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places. Answer to each question will be evaluated according to the following marking scheme: Full Marks : + 3 If ONLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. 9. The decomposition reaction D 2N 2O 5(g) ¾® 2N 2O 4 (g) + O 2 (g) is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After Y ´ 10 3 s, the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is 5 ´ 10 -4 s -1 , assuming ideal 11. Total number of hydroxyl groups present in a molecule of the major product P is ……… (i) H2, Pd-BaSO4, quinoline (ii) Dil. KMnO4, (excess), 273 K P gas behaviour, the value of Y is ……… 10. The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm -3 , then molarity of urea solution is …… (Given data : Molar masses of urea and water are 60 g mol -1 and 18 g mol -1 , respectively) 12. Total number of cis N ¾ Mn ¾ Cl bond angles (that is Mn ¾ N and Mn ¾ Cl bonds in cis positions) present in a molecule of cis [Mn(en)2 Cl 2 ] complex is ……… (en = NH 2CH 2CH 2NH 2 ) 13. The amount of water produced (in g) in the oxidation of 1 mole of rhombic sulphur by conc. HNO 3 to a compound with the highest oxidation state of sulphur is …… (Given data : Molar mass of water = 18 g mol -1 ) 14. Total number of isomers considering both structural and stereoisomers of cyclic ethers with the molecular formula C4H 8O is ……… Section 3 (Maximum Marks : 12) l l l l l l This section contains TWO (02) List-Match sets. Each List-Match set has TWO (02) Multiple Choice Questions. Each List-Match set has two lists : List-I and List-II List-I has Four entries (I), (II), (III) and (IV) and List-II has Six entries (P), (Q), (R), (S), (T) and (U). FOUR options are given in each Multiple Choice Question based on List-I and List-II and ONLY ONE of these four options satisfies the condition asked in the Multiple Choice Questions. Answer to each question will be evaluated according to the following marking scheme : Full Marks : + 3 If ONLY the option corresponding to the correct combination is chosen: Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : - 1 In all other cases. Answer the following by appropriately matching the lists based on the information given in the paragraph. 15. Consider the Bohr’s model of a one-electron atom where the electron moves around the nucleus. In the following List-I contains some quantities for the nth orbit of the atom and List-II contains options showing how they depend on n. List-I List-II (I) Radius of the n th orbit (P) µ n -2 (II) Angular momentum of the electron in the nth orbit (Q) µ n -1 (III) Kinetic energy of the electron in the nth orbit (R) µ n0 (IV) Potential energy of the electron in the nth orbit (S) µ n1 (T) µ n2 (U) µ n1/ 2 (a) (II), (R) (c) (I), (T) (b) (I), (P) (d) (II), (Q) 17. List-I includes starting materials and reagents of selected chemical reactions. List-II gives structures of compounds that may be formed as intermediate products and/or final products from the reactions of List-I. List-I CN (I) O O (II) CO2H Cl CO2CH3 (b) (III), (S) (d) (IV), (Q) 16. Consider the Bohr’s model of a one-electron atom (IV) CO2Me Radius of the nth orbit (P) µ n -2 (II) Angular momentum of the electron in the nth orbit (Q) µ n -1 (iii) NaBH4 (iv) Conc. H2SO4 CHO CO2H OH (Q) OH (iii) NaBH4 (iv) Conc. H2SO4 (i) KCN (ii) H3O+,D (R) O (iii) LiAIH4 (iv) Conc. H2SO4 (iii) LiAIH4 OH (S) CO2H (T) CO2H CO2H List-II (I) (P) CO2Me (iv) Conc. H2SO4 where the electron moves around the nucleus. In the following List-I contains some quantities for the nth orbit of the atom and List-II contains options showing how they depend on n. List-I List-II (i) DIBAL-H (ii) Dil. HCl (i) O3 (ii) Zn, H2O (III) Which of the following options has the correct combination considering List-I and List-II? (a) (III), (P) (c) (IV), (U) Which of the following options has the correct combination considering List-I and List-II? (U) O O 0 (III) Kinetic energy of the electron in the nth (R) orbit µn (IV) Potential energy of the electron in the nth orbit (S) µ n1 (T) µ n2 (U) µ n1/ 2 Which of the following options has correct combination considering List-I and List-II? (a) (III), (S), (R) (b) (IV), (Q), (R) (c) (III), (T), (U) (d) (IV), (Q), (U) 18. List-I includes starting materials and reagents of List-I selected chemical reactions. List-II gives structures of compounds that may be formed as intermediate products and/or final products from the reactions of List-I. List-I CN (I) O O (II) (III) Cl CO2CH3 (IV) (P) CO2Me (iv) Conc. H2SO4 CO2H (T) CHO (iii) NaBH4 (iv) Conc. H2SO4 CO2H CO2H CO2H (U) O OH (Q) O OH (iii) NaBH4 (iv) Conc. H2SO4 (i) KCN (ii) H3O+,D OH (S) (iii) LiAIH4 List-II (i) DIBAL-H (ii) Dil. HCl (i) O3 (ii) Zn, H2O CO2H List-II CO2Me Which of the following options has correct combination considering List-I and List-II? (R) O (iii) LiAIH4 (iv) Conc. H2SO4 (a) (II), (P), (S), (U) (c) (II), (P), (S), (T) (b) (I), (Q), (T), (U) (d) (I), (S), (Q), (R) Answer with Explanations Paper 1 D Na 2B 4O 7 ¾® B 2O 3 + 2NaBO 2 1. (b) Acidic nature depends upon nature of electron withdrawing group and electronegativity. Electronegativity further depends on % s character. Higher the s-character, greater will be the electronegativity and hence tendency to loose H increases thus acidic character also increases. O O (I) H H O OH 2 sp -hybridisation (Resonance effect) (pKa = 4.5) Boric anhydride is non-volalite. When it react with Cr(III) salt then deep green complex is formed. 2Cr 3+ + 3B 2O 3 ¾® 2Cr(BO 2)3 Deep green H sp 2 -hybridisation (30-33% s character) (pKa = 4.3) (III) MeO OH (IV) H3C Sodium glassy bead OH (II) OH sp- hybridisation (50% s character) (pKa = 1.86) Boric anhydride 1 4442metaborat 4443e O 3 sp -hybridisation (25% s-character) (pKa = 4.8) Hence, acidic order I > II > III > IV. II is more acidic than III since electron donating group (¾OCH 3) is attached to benzene ring in III which decreases the acidic character. On the other hand, pK a value also determined acidic nature, lower pK a value gives maximum acidic character. Hence, option (b) is correct. 2. (c) Borax bead test is performed only for coloured salt. Borax (sodium pyroborate), Na 2B 4O 7 ×10H 2O on heating gets fused and lose water of crystallisation. It swells up into fluffy white porous mass which melts into a colourless liquid which later form a clear transparent glassy bead consisting of boric anhydride and sodium metaborate. D Na 2B 4O 7 ×10H 2O ¾® Na 2B 4O 7 +10H 2O ­ Hence, option (c) is correct. 3. (d) ZnCO 3-Calamine (zinc ore) CuCO 3 × Cu(OH)2-Malachite (copper ore) Fe3O 4 -Magnetite (iron ore) Na 3AlF6-Cryolite (aluminium ore) Thus, option (d) is correct. 4. (b) Key Idea The aqueous solution of ionic surfactant, i.e. – + sodium stearate (C17H 35CO O N a) acts as a strong univalent type of electrolyte in the concentration range below the CMC and the linear function of dependence of L m on C has a small negative slope. CMC of Lm on ÖC has a small negative slope. Lm ¾ ÖC At normal or low concentration, sodium stearate [CH 3(CH 2)16COO -Na + ] behaves as strong electrolyte and for strong electrolyte, molar conductance (Lm) decreases with increase in concentration. Above particular concentration, sodium stearate forms aggregates known as micelles. The concentration is called as CMC. Since, number of ions decreases and hence Lm also decreases. Hence, option (b) is correct. 5. (c, d) The standard enthalpy of formation is defined as standard enthalpy change for formation of 1 mole of a substance from its elements, present in their most stable state of aggregation. 3 O 2(g) ¾® O 3(g) ; 2 1 S8 (s) + O 2(g) ¾® SO 2(g) 8 In the above two reactions standard enthalpy of reaction is equal to standard enthalpy of formation. 6. (a, d) – – SnCl2 + Cl SnCl3 Tin chloride (Q) Cl (X) Sn Cl Cl sp3 (pyramidal) SnCl -3 has (3s + 1lp) and exist in pyramidal structure. Me N Me Me SnCl2 × NMe3 SnCl2+Me3N (3°amine) (Q) (Y) 8. (a, b, d) The explanation of given statements are as follows: (a) Urms is inversely proportional to the square root of its molecular mass. 3RT Urms = M Hence, option (a) is correct. (b) When temperature is increased four times then Urms become doubled. 3R Urms = ´ 4T M 3RT Urms = 2 ´ M Hence, option (b) is correct. (c) and (d) E av is directly proportional to temperature but does not depends on its molecular mass at a given temperature as 3 E av = KT. If temperature raised four times than E av becomes 2 four time multiple. Thus, option (c) is incorrect and option (d) is correct. 9. (c, d) The given road map problem is Sn CH3MgBr Cl +4 SnCl2 + 2CuCl2 H Conc. HCl +1 SnCl4 + 2CuCl (Q) Cl OH (Q) Major (S) Cl CH3 CH3 Br (i) H2,Ni (Reduction) Cl Sn + H2O (ii) Br2,hn (free radical substitution) major (T) Cl Br + Mg –H2O (Z) Oxidation Z is oxidised product and oxidation state of Sn is +4 in Z compound. Structure of SnCl 4 (Z) is Hydrolysis CH3 OH CH3 Cl Reduction +2 H2O (C6H10O) Y complex has coordinate bond in between nitrogen and Sn metal. +2 H—OH CH3 O MgBr O Cl 20% H3PO4, 360K Dehydration of alcohol major (R) CH3 Thus, options (a, d) are correct. Br 7. (a,b,c) Key Idea The lose of one a-particle will decrease the HBr mass number by 4 and atomic number by 2. On the other hand, loss of b-particle will increase the atomic number by 1. major (U) Hence, options (c, d) are correct. In decay sequence, 92U 234 + 2He4 (or a) 90Th 238 X1 particle 234 + (b– or –1e0) 92U Z234 is isotope of uranium Benzoyl peroxide (anti-Markovnikov's rule) 91Pa 234 + (b– or –1e0) 10. (a, c, d) The explanation of given statements are as follows: (a) Two six membered cyclic hemiacetal form of D-(+)- glucose are called anomers. X2 particle X3 particle CH2OH O H 230 90Th H OH + 2He (or a) 4 X4 particle X1 particle will deflect towards negatively charged plate due to presence of positive charge on a- particles. Hence, options (a, b, c) are correct. OH H CH2OH OH H H OH a-D(+)-glucopyranose Both are anomers. O H H OH OH H H H OH OH b-D(+)-glucopyranose (b) Oxidation of glucose in presence of Br2 water gives gluconic acid. CHO The molecules which gives permanent dipole moment are polar in nature. Cl COOH H HO OH H H H HO H Br2 OH Water H H O (c) Monosaccharides can not be hydrolysed into polyhydroxy aldehydes and ketones. F Invertase C6 H12O 6 + D-glucose or dextrorotatory K2MnO4 (aq) (W) (W) potassium manganate 92K (aq) + MnO 4 2– C (aq) 3 Mn O O O sp- hybridisation, tetrahedral (manganate ion) Green coloured complex é +e (Z) and MnO -4 both ions form p-bonding between MnO 24 (Y) Cl m¹0 (polar) m¹0 (polar) B Cl , C F O , F m=0 (non-polar) (polar) O F , F Xe O F m¹0 (polar) F F m»0 (non-polar) Thus, options (a, c) are correct. N ºº N ¾® O , (N2O) ( Z) N¾N OH HO O O (N2O4) , O (H2S2O3) O O S H S O¾O HO O (H2S2O8) OH 12. (a,c) Key Idea Dipole moment of a bond depends on the H H H | B H H B B , O H p-orbitals of oxygen and d-orbitals of manganese. Thus, options (a, c, d) are correct. difference in the electronegativities of bonded atoms. More is the difference in the electronegativities, greater will be the dipole moment. Also, For symmetrical molecule, m = 0 For unsymmetrical molecule, m ¹ 0 O O S - In acidic medium, Y undergoes disproportionation reaction. 3MnO 24- (aq) + 4H+ ¾® 2MnO -4 + MnO 2 + 2H 2O (Y) Cl , P containing covalent bond between two atoms. é Q m¹0 (polar) H m¹0 MnO -4 ions gives charge transfer spectrum in which a fraction of electronic charge is transferred between the molecular entities. Electrolytic MnO 24- ¾¾¾® MnO -4 oxidation (Y) ( Z) O H H H S O sp3, tetrahedral Mn (purple coloured complex ion) O O O é O 13. (4.00) N 2O, N 2O 4 , H 2S2O 3 and H 2S2O 8 molecules are ( X) K+ + MnO4(Z) O , F H K 2MnO 4 + H 2O ¾¾¾¾¾¾¾®H 2 + KOH + KMnO 4 D Cl N , F Electrolytic oxidation ( W) m 0 (non-polar) N Cl Cl H MnO 24 ion has one unpaired electrons, therefore it gives d-d transition to form green colour. Y complex has paramagnetic nature due to presence of one unpaired electron. In aqueous solution, KMnO4(aq) C é é é – O B Cl Cl m¹0 (polar) (Y) Cl , O Cl , Cl mnet» 0 (non-polar) mnet 0 (non-polar) H m2 Be Cl m2 O L-fructose or laevorotatory D 11. (a, c, d) MnO 2 + 2KOH + O 2 ¾® K 2MnO 4 + H 2O + m1 , F m1 C6H12O 6 Hence, options (a, c, d) are correct. 1 2 m¹0 (polar) F m¹0 (polar) (d) Hydrolysis of sucrose gives D-glucose and L-fructose. H H Br Gluconic acid C12H 22O11 + H 2O ¾¾¾¾® , m¹0 (polar) F F CH2OH D-(+)-glucose , m¹0 (polar) OH CH2OH Se O OH H OH S OH H H (B2H6) H N N B B N H (B3N3H6) B 2H 6 and B 3N 6H 6 have polar bond, but do not have same kind of atom. 14. (1.02) Key Idea First calculate, molar mass of solute using the p ° - ps nsolute formula, and then calculate DTf = p° nsolute + nsolvent by applying the formula; DTf = K f ´ m. When 0.5 g of non-volatile solute dissolve into 39 gm of benzene then relative lowering of vapour pressure occurs. Hence, vapour pressure decreases from 650 mmHg to 640 mmHg. Given, vapour pressure of solvent (p°) = 650 mmHg Vapour pressure of solution (ps) = 640 mmHg Weight of non-volatile solute = 0.5 g Weight of solvent (benzene) = 39 g From relative lowering of vapour pressure, nsolute p ° - ps = xSolute = p° nsolute + nsolvent 0.5 650 - 640 molar mass = 0.5 39 650 + molar mass 78 0.5 10 molar mass = 0.5 650 + 0.5 molar mass 0.5 + 0.5 ´ molar mass = 65 ´ 0.5 \ Molar mass of solute = 64 g From molal depression of freezing point, DTf = K f ´ molality K f ´ wsolute = (MW)solute ´ wsolvent 0.5 ´ 1000 64 ´ 39 Fe2+ (aq) + S2- (aq) 1 At initial concentration (Before mixing) At initial concentration (After mixing) At equilibrium (Rate)1 [0.2]x [01 . ]y [01 . ]z 6 ´10-5 = = x y z (Rate)2 [0.2] [0.2] [01 .] 6 ´10-5 – 0.03 M 0.1 M – 0.03- X 0.1 – 0.03 = 0.07 X = Y ´10-17 = 8.9 ´10-17 Y = 8.9 Given, \ 17. (19) XeF4 reacts with O 2F2 to form XeF6 × O 2F2 is fluoronating reagent. 143 K XeF4 + O 2F2 ¾¾® XeF6 + O 2 (Y) The structore of XeF6 is F F F Xe F F F Y compound (XeF6) has 3 lone pair in each fluorine and one lone pair in xenon. Hence, total number of lone pairs electrons is 19. 18. (4) Scheme -1 NH2 Þ y=0 (Rate)1 [0.2]x [01 . ]y [01 . ]z 6 ´10-5 = = x y z (Rate)3 [0.2] [01 1.2 ´10-4 . ] [0.2] NH2 Br H2O Aniline (P) Br 273K (diazotisation of aniline ) Br 2,4,6-tribromoaniline + – N 2 Cl Br Br + N2 From exp-Ist, Rate = 6.0 ´ 10-5 mol dm -3 s -1 – Cl is displace by CN–ion + (iv) H3O dm -3 [A] = 015mol . [B] = 0.25 mol dm -3 [C] = 015 . mol dm -3 ¾¾¾® k = 3 ´ 10-3 (complete hydrolysis of ¾ CN group which convert into ¾COOH) Br 2, 4, 6 tribromo diazonium chloride salt O C ¾Cl COOH Br (v) SOCl2, Br Br Br Pyridine ¾¾¾¾® . ]1 [0.25]0 [015 . ]1 = 3 ´ 10-3 ´ 015 . ´ 015 . \Rate = (3 ´ 10-3) ´ [015 Rate = 6.75 ´ 10-5 mol dm -3 s -1 Thus, Y = 6.75 Br Br (iii) CuCN/KCN Br 6.0 ´ 10-5 = k[0.2]1 [01 . ]1 (ii) NaNO2 +HCl ¾¾¾¾¾¾® CN Þ x =1 So, rate = k[A]1 [C]1 Br (i) Br2(excess) Þ z =1 (Rate)1 [0.2]x [01 . ]y [01 . ]z 6 ´10-5 = = x y (Rate)4 [0.3] [01 . ] [01 . ]2 9 ´10-5 Given, 0.2 M – [Here, K c >>10 , thus limiting reagent will be consumed almost completely, 0.03 - X = 0 \ X = 0.03] From equilibrium constant, [FeS] KC = [Fe2+][S2- ] 1 [For FeS(s) = 1 mol L -1 ] KC = X ´ 0.07 (Pure solid) 1 17 1.6 ´ 10 = X ´ 0.07 1 X= = 8.9 ´ 10-17 17 1.6 ´ 10 ´ 0.07 Þ DTf = 1.02K 15. (6.75) Rate = k[A]x [B]y [C]z FeS (s) 0.06 M 3 ® DTf = 512 . ´ 16. (8.9) Given, equilibrium constant (K c) at 298 K = 1.6 ´ 1017 Br Major (Q) Scheme-2 SO3Na SO3H (R ) + (ii) NaOH, D (i) Oleum ¾¾¾¾® (iii) H Hydrolysis ¾¾¾¾® H2S2O7 (H2SO4+ SO3) (Sulphonation) OH OH OH Br2 + NaHSO3 ¾¾¾¾® CS2, 273K (Bromination) ¾¾¾¾¾¾¾® Br + minor 2- bromophenol Br major (S) 4- bromophenol Scheme-3 O Br O–Na+ OH (i) NaOH, D O C Br (Q) Br ¾¾¾® Br + NaCl Br Br Br O C¾Cl Br Br (T) (Major) (S) T compound has total number of Br atom =4 Paper 2 1. (a, b) C C CH2 CHO C (i) Hg2+, dil. H2SO4 Hydration of Alkyne C CH2 CH2 O C (¾CHO converts into COO–) O CH2 CHO O CH2 CH2 CH2 (iii) Zn – Hg + Conc. HCl C ¾ reduce to ¾ CH2 (ii) AlCl3 CH2 CH3O (Q) CH2 (i) SOCl2+Pyridine (Clemmensen reduction) CH3O H Cl COOH O C CH2 O CH2 CH2 CH3O CH2 CH2 CH2 Zn-Hg/Conc. HCl Clemmensen reduction CH3O (S) (a) CH2 CH3O CH3O 2. (c, d) CHO Keto–enol Tautomerism (ii) AgNO3+NH4OH (Tollen's reagent) O– C CH2 OH CH3O CH3O CH C CH2 CH2 Electrophilic substitution reaction O (R) Key Idea An aromatic compound must be cyclic and planar. It must follow ( 4 n + 2 ) e - rule and have the conjugated system in it. Br OEt NaOEt Dimerise + Substitution product (Non-aromatic) Ellimination product (b) Benzene react with Cl 2 (excess) in presence of UV light and 500 K of temperature to form benzene hexachloride (non-aromatic). Cl + Cl2 (excess) Cl Cl Cl Cl UV 500 K Cl (Non-aromatic) (c) CH3 (i) Alc. KOH Br (ii) NaNH2 CH3C CH Au+ HNO 3 +4HCl ¾® AuCl –4 + H 3O + + NO + H 2O 873 K (Trimerisation) Br +3 0 (iii) Red hot iron tube Oxidation CH3 Hence, option (c) is correct. CH3 CH3 (Aromatic) H 6. (a, d) The explanation of given statements are as follows: (a) Teflon is prepared by heating tetrafluoroethene in presence of persulphate catalyst at higher pressure. s Na+ (d) (d) Reaction of gold with aqua-regia produces NO gas in absence of air. Hence, option (d) is incorrect. nCF2 ==CF2 NaOMe –MeOH Persulphate catalyst (Aromatic ion) Thus, (c) and (d) options are correct. 3. (a, c) Given, ground state energy of hydrogen atom = - 136 . eV Thus, option (a) is correct. (b) Natural rubber is polyisoprene containing cis alkene units. 136 . ´ Z2 eV n2 - 136 . ´ (2)2 136 . ´4 - 34 . eV = Þ n= Þ n= 4 34 . n2 Given, azimuthal quantum number () l = 2(d – subshell Magnetic quantum number (m) = 0 \ Angular nodes () l =2 Radial node = n - l - 1 = 4 - 2 - 1 = 1 nl = 4d state Hence, options (a), (c) are correct. nCH2 C Energy of He + , E = - released and ZnSO 4 is obtained. Zn + 2H 2 SO4 ¾® ZnSO4 + SO2 ­ + 2H 2 O (Hot + Conc.) (G) (R) (X) R(SO2) molecule is V-Shaped H cis-ployisoprene unit (natural rubber) Isoprene OH OH OH O O O OH O 1 O OH O OH OH (1, 4, b -linkage) OH O OH O OH , 4 OH n OH O (cellulose) O O NH D, H2O O Thus, option (b) is correct. When Zn is react with conc. NaOH then H 2 gas is evolved and Na2ZnO2 is obtained. Zn + 2NaOH (conc.) ¾® Na2ZnO2 + H 2 ­ ( Q) In ground state, H—H (Q) (bond order = 1) Thus, option (c) is correct. The oxidation state of Zn in T(Na 2ZnO 2) is +2 Thus, option (a) is incorrect. ZnSO4 + H 2 S + NH 4 OH ¾® ZnS¯ + 2H 2 O + (NH 4 )2 SO4 ( X) C Thus, option (b) is incorrect. (c) Cellulose has only b-D-glucose units that are joined together by glycosidic linkages as shown in the following structure: C ( Z) C CH3 Thus, option (c) is incorrect. (d) Nylon-6 has amide linkages. S ( T) CH2 CH CH3 4. (b, c, d) When Zn react with hot conc. H 2 SO4 then SO2 is ( G) CH2 n CH2 Energy of He + = - 34 . eV, Z = 2 O CF2 CF2 Thermoplastic Polymer (PTFE) (Y) ZnS (Z) compound is dirty white coloured. Thus, option (d) is correct. 5. (a, b, c) The explanation of given statements are as follows: (a) Aqua-regia is prepared by mixing conc. HCl and conc. HNO 3 in 3:1 (v/v) ratio and is used in oxidation of gold and platinum. Hence, option (a) is correct. (b) Yellow colour of aqua-regia is due to its decomposition into NOCl (orange yellow) and Cl 2 (greenish yellow). Hence, option (b) is correct. (c) When gold reacts with aqua-regia then it produces AuCl -4 anion complex in which Au has +3 oxidation state. [H2N(CH2)5COOH] –nH2O D C (CH2)5 NH Nylon-6 n Caprolactam Thus, option (d) is correct. 7. (a, b, c) Cyanide process of gold extraction involves leaching out gold from its ore with CN - in the presence of O 2 (Q) in water to form [Au(CN)2 ]- (R). When [Au(CN)2 ]- reacts with Zn (T), it froms [Zn(CN)4 ]2- (Z) and Au. The corresponding reactions are as follows : H O + O2 ( Q) 2 4 Au (s) + 8 CN - (aq) ¾¾¾¾® 4[Au(CN)2 ]- + 4OH - (aq) (R) 2[Au(CN)2]- (aq) + Zn(s) ¾® [Zn(CN)4 ]2- (aq) + 2 Au(s) ( R) ( T) Hence, options (a, b, c) are correct. 8. (a, c) The given reactions takes place as follows: (1) Cl (a) CH3 Zn, dil. HCl CH3 CH2 (Propane) CH3 Br Br (b) CH 3 11. (6) Zn CH3 D CH3 CH2 + ZnBr2 CH3 Electrolysis CH2 Quinoline (reduction) (Triple bond converts into double bond) CH3+ CO2 (Propane) COONa (d) CH3 (i) H2, Pd–BaSO4 (Propene) NaOH, CaO, D Electrolysis (c) CH (Elimination reaction) CH3 (CH2)4 CH3 n-hexane COONa CH2 Thus, options (a, c) are correct. (ii) Dil. KMnO 4 (excess) 273K 9. (2.3) At constant V, T H 2N 2 O5 (g) ¾® 2N 2 O4 (g) + O2(g) 0 2p 0 p pTotal = 1 - 2p + 2p + p 1. 4 = 1 + p p = 0.45 atm According to first order reaction, pi 2303 . k= log t pi - 2p pi = 1atm (given) 2p = 2 ´ 0.45 = 0.9 atm On substituting the values in above equation, 1 2k × t = 2.303log 1 - 0.9 1 2 ´ 5 ´ 10-4 ´ y ´ 103 = 2.303log 01 . y = 2.303 = 2.3 Note Unit of rate constant (k), i.e. s-1 represents that it is a first order reaction. 10. (2.98 mole) Key Idea Molarity ( M) = C Number of moles of solute ´ 1000 Volume of solution (in mL) Mass Also, volume = Density Given, mole fraction of urea (c urea) = 0.05 Mass of water = 900g Density = 1.2 g/cm 3 nurea 900 [Q Moles of water = c urea = = 50] nurea + 50 18 nurea 0.05 = Þ 19nurea = 50 nurea + 50 nurea = 2.6315 moles wurea = nurea ´ (M × wt)urea = (2. 6315 ´ 60)g 2.6315 ´ 60 + 900 é Mass of solution ù V= Q Density = êë 1.2 Volume of solution úû = 881.57 mL Now, molarity 1000 = Number of moles of solute ´ Volume of solution (mL) 2.6315 ´ 1000 = = 2.98 M 881.57 H OH D At initial t = 0 1 t = Y ´103 sec 1 - 2p CH2 C H H C C OH OH CH2 OH OH OH (P) (Major product) Compound (P) has total number of hydroxyl groups = 6 12. (6) The structure of cis-[Mn(en)2Cl 2] complex is Cl (a) Cl N1 (b ) Mn Cl (b) Mn en en Cl (a) N2 N N4 3 Bond angles (Mn—N and Mn—Cl bond in cis positions) Cl (a) —— Mn —— N(1) Cl (a) —— Mn —— N(2) Cl (a) —— Mn —— N(4) Cl (b) —— Mn —— N1 Cl (b) —— Mn —— N3 Cl (b) —— Mn —— N4 Number of cis Cl—Mn—N = 6 13.(288) Key Idea Rhombic sulphur (S 8 ) gets oxidised into sulphuric acid and water, NO 2 gas is released on reaction with conc. HNO 3. When rhombic sulphur (S8 ) is oxidised by conc. HNO3 then H 2 SO4 is obtained and NO 2 gas is released. S8 + 48HNO3 ¾® 8H 2 SO4 + 48NO2 + 16 H 2O 1 mole of rhombic sulphur produces = 16 moles of H 2 O \ Mass of water = 16 ´ 18 (molar mass of H 2 O) = 288 g 14. (10.0) The structure of cyclic ether with molecular formula, C4H 8O are as follows: (chiral carbon) (1) O (2) * , O R isomer = 2 isomer S isomer O (3) O , (4) * (1-chiral carbon) R isomer S isomer = 2 isomer O (5) * O (6) , * O R, R isomer R, S isomer = 3 isomer S, S isomer C (2-chiral carbon) OMe (i) LiAlH4 (IV) C Total number of isomers of cyclic ether with molecular formula, C4 H 8 O are 10. CH2 (Q) O 15. (a) (III) Kinetic energy of the electron in nth orbit, Z2 K.E. = + 136 . ´ 2 n 1 or K.E. µ 2 or K.E. µ n-2 n From list-II, correct match is (III P). (IV) Potential energy of the electron in the nth orbit, Z2 P.E. = - 2 ´ 136 . ´ 2 n 1 P.E. µ 2 n P.E. µ n-2 O O 18. (a) CH2 CN (I) O CH2OH CH2 OH (iv) Conc. H2SO4 (dehydration) Hence, correct match of (I) are (Q, R) O LiAlH4 (iv) conc. H2SO4 D CH2 (i) O3, (ii) Zn + H2O (II) C H (Ozonolysis) COOH (P) COOH (iii) NaBH4 C OCH3 (iv) Conc. H2SO4 O CH2 (ii) H3O+ D (Complete hydrolysis of —CN give —COOH and ester also get hydrolysed into —COOH and alcohol ) H 2O + Hence, correct match of (III) are T , Q, R. O (R) (U) OH –H2O C COOH CH2 CH2 OH O (R) O OH C CN (i) KCN O (Q) CH H (iii) NaBH4 O O CH2 OH –H2O O (Q ) Nucleophilic substitution reaction OCH3 (Cl– replaces by CN– ion) (T) OH O (S) C H OH (ii) Dil. HCl (reduction) 17. (b) CH3OH + C (i) DIBAL–H CH From list-II, correct match is (I, T) (II) Angular momentum of the electron, nh or mvr µ n mvr = 2p From list-II, correct match (II, S) Hence, correct matching from list-I and list-II on the basis of given option is (I, T). C Dehydration Hence, correct matching from list-I and list II on the basis of given option is (IV), Q, R. n2 r = 0.529 ´ Z r µ n2 Cl (ii) conc. H2SO4 Hence, correct match of IV is Q, R. 16. (c) (I) Radius of the nth orbit, (III) + H2O CH2OH (R) From List II, correct match is (IV P). Hence, correct matching from list-I and list-II on the basis of given option is (III, P). Here, CH2OH OMe O OH O Hence, correct match of II is (P , S ,U). Hence, correct matching from list-I and list-II on the basis of given option is (II), P , S , U.